[NOI2014]动物园

题解:

先求出可重复的nex[]和num[].然后再次做kmp时,对于一个位置i一直跳到失配位置j,且(j<<1)<=i,所以此时重叠的num[i]=此时的num[j]+1.

#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<queue>
#include<vector>
#include<stack>
#define MAXN 1000010
#define RG register
#define LL long long int
using namespace std;
const int INF=1e9;
const int mod=1e9+7;
int T;
int n;
char s[MAXN];
int nex[MAXN],num[MAXN];
LL ans;
void work()
{
  int j=0;ans=1;num[1]=1;
  for(int i=2;i<=n;i++)
    {
      while(j>0&&s[j+1]!=s[i]) j=nex[j];
      if(s[j+1]==s[i]) j++;
      nex[i]=j;num[i]=num[j]+1;
    }
  j=0;
  for(int i=2;i<=n;i++)
    {
      while(j>0&&s[j+1]!=s[i]) j=nex[j];
      if(s[j+1]==s[i]) j++;
      while((j<<1)>i&&nex[j]) j=nex[j];
      ans=ans*(num[j]+1)%mod;
    }
}
int main()
{
  freopen("1.in","r",stdin);
  scanf("%d",&T);
  while(T--)
    {
      scanf("%s",s+1);n=strlen(s+1);
      work();
      printf("%lld
",ans);
    }
  return 0;
}